What would be the final ph if 0.0100 moles of solid naoh were added to 100ml of a buffer solution containing 0.600 molar formic acid (ionization constant = 1.8x10-4) and 0.300 m sodium formate?

Moles of formic acid = 0.6 M * 0.1L = 0.06 m
moles of sodium formate = 0.3 M * 0.1L = 0.03 m
According to the reaction equation:
 
                     HCOOH(aq) + NaOH(s) ↔ HCOO- (aq) + H2O(l)
intial             0.06                  0.01               0.03
change       - 0.01                 -0.01               +0.01
final             0.05                    0                   0.04

So when :
PH =PKa + ㏒(strong base/weak acid)
when we have Ka =1.8x10^-4 and sodium formate is the strong base and formic acid is the weak acid.
PH = -㏒ Ka + ㏒(0.04/0.05)
      = -㏒1.8x10^-4 + ㏒ (0.04/0.05)
 ∴PH  =3.65

Well to my expert knowledge and my exquisite calculations i think the the negative ion # 412 is the key to sucess